Question

A body of mass $10kg$ slides along a rough horizontal surface. The coefficient of friction is $1/\sqrt{3}$. Taking $g=10m/s^2$, the least force which acts at an angle of $30^{\circ }$ to the horizontal is

• A. $25N$
• B. $100N$
• C. $50N$
• D. $\frac{50}{\sqrt{3}}N$

Answer 1

Answer: (C)

Let $P$ force is acting at an angle $30^{\circ }$ with the horizontal.
For the condition of motion $F=\mu R$
$P\cos 30^{\circ }=\mu \left(mg-P\sin 30^{\circ }\right)$
$\Rightarrow P\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}}\left(100-P\frac{1}{2}\right)$
$\Rightarrow \frac{3P}{2}=\left(100-\frac{P}{2}\right)$
$\Rightarrow 2P=100$
$\Rightarrow P=50N$